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Let's solve each of the given limits one by one.

### 11. \(\lim_{{x \to 3}} (2x^2 + 4x + 1)\)
We can directly substitute \(x = 3\) into the quadratic expression:
\[
2(3)^2 + 4(3) + 1 = 2(9) + 12 + 1 = 18 + 12 + 1 = 31
\]
So, the limit is \(\boxed{31}\).

### 12. \(\lim_{{x \to 1}} (3x^3 - 2x^2 + 4)\)
Direct substitution of \(x = 1\):
\[
3(1)^3 - 2(1)^2 + 4 = 3 - 2 + 4 = 5
\]
The limit is \(\boxed{5}\).

### 13. \(\lim_{{x \to 3}} \sqrt{x + 1}\)
Direct substitution of \(x = 3\):
\[
\sqrt{3 + 1} = \sqrt{4} = 2
\]
The limit is \(\boxed{2}\).

### 14. \(\lim_{{x \to 4}} \frac{3}{x + 4}\)
Substitute \(x = 4\):
\[
\frac{3}{4 + 4} = \frac{3}{8}
\]
The limit is \(\boxed{\frac{3}{8}}\).

### 15. \(\lim_{{x \to 4}} (x + 3)^2\)
Substitute \(x = 4\):
\[
(4 + 3)^2 = 7^2 = 49
\]
The limit is \(\boxed{49}\).

### 16. \(\lim_{{x \to 0}} (2x - 1)^3\)
Substitute \(x = 0\):
\[
(2(0) - 1)^3 = (-1)^3 = -1
\]
The limit is \(\boxed{-1}\).

### 17. \(\lim_{{x \to 2}} \frac{1}{x}\)
Substitute \(x = 2\):
\[
\frac{1}{2}
\]
The limit is \(\boxed{\frac{1}{2}}\).

### 18. \(\lim_{{x \to -3}} \frac{2}{x + 2}\)
Substitute \(x = -3\):
\[
\frac{2}{-3 + 2} = \frac{2}{-1} = -2
\]
The limit is \(\boxed{-2}\).

### 19. \(\lim_{{x \to 1}} \frac{x}{x^2 + 4}\)
Substitute \(x = 1\):
\[
\frac{1}{1^2 + 4} = \frac{1}{1 + 4} = \frac{1}{5}
\]
The limit is \(\boxed{\frac{1}{5}}\).

### 20. \(\lim_{{x \to -2}} \frac{2x - 3}{x + 5}\)
Substitute \(x = -2\):
\[
\frac{2(-2) - 3}{-2 + 5} = \frac{-4 - 3}{3} = \frac{-7}{3}
\]
The limit is \(\boxed{-\frac{7}{3}}\).

### 21. \(\lim_{{x \to 7}} \frac{3x}{\sqrt{x + 2}}\)
Substitute \(x = 7\):
\[
\frac{3(7)}{\sqrt{7 + 2}} = \frac{21}{\sqrt{9}} = \frac{21}{3} = 7
\]
The limit is \(\boxed{7}\).

### 22. \(\lim_{{x \to 2}} \frac{\sqrt{x + 2}}{x - 4}\)
Substitute \(x = 2\):
\[
\frac{\sqrt{2 + 2}}{2 - 4} = \frac{\sqrt{4}}{-2} = \frac{2}{-2} = -1
\]
The limit is \(\boxed{-1}\).

Would you like a more detailed explanation for any of these solutions or have any specific questions?

Here are some related questions you can explore:
1. What happens if the limit approaches a point where the function is undefined?
2. How can we compute limits at infinity for rational functions?
3. What are the techniques for solving indeterminate forms like \(\frac{0}{0}\)?
4. How do you apply L'Hopital's Rule to limits?
5. What is the concept of continuity in relation to limits?

**Tip:** Always attempt to substitute the point into the limit first to see if direct substitution works. If not, explore other techniques like factoring or L'Hopital's Rule.

Evaluate the limits for the given mathematical expressions.

Learn how to solve a variety of limit problems, including limits of polynomial and rational functions, with detailed step-by-step solutions. Suitable for students in Grades 10-12.

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